Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the (homework) tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post.
–
Cameron BuieNov 13 '12 at 20:33

2 Answers
2

Take $K=\mathbb{Z}_2(X)$ and $L=K(\sqrt[3]{X})$. Then the polynomial $f(T)=T^3-X\in K[T]$ is irreducible (because it has no roots in $K$), has one root in $L$, namely $\sqrt[3]{X}$, and I leave you the pleasure to prove that this is the only root of $f$ in $L$.

Here's a hint/general outline: We know that finite extensions of finite fields are normal, so we're going to need something a little more complicated. The simplest example here would be to take an extension of $\mathbb{F}_p(T)$, where $T$ is transcendental.

Over the rationals, the canonical example is to adjoin an $n^{\text th}$ root ($n>2$) of some integer $a$, by adding a root of the equation $X^n-a=0$. In short, this will fail to be normal because $\mathbb{Q}$ does not include the $n^{\text th}$ roots of $1$.

Try something similar with taking an $n^{\text th}$ root of $T$, where the $n^{\text th}$ roots of $1$ are not in $\mathbb{F}_p$.